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More Number Theory Proofs Rosen 1.5, 3.1

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Prove or Disprove If m and n are even integers, then mn is divisible by 4. The sum of two odd integers is odd. The sum of two odd integers is even. If n is a positive integer, then n is even iff 3n 2 +8 is even. n 2 + n + 1 is a prime number whenever n is a positive integer. n 2 + n + 1 is a prime number whenever n is a prime number. |x| + |y| |x + y| when x,y R. 3 is irrational.

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If m and n are even integers, then mn is divisible by 4. Proof: m and n are even means that there exists integers a and b such that m =2a and n = 2b Therefore mn = 4ab. Since ab is an integer, mn is 4 times an integer so it is divisible by 4.

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The sum of two odd integers is odd. This is false. A counter example is 1+3 = 4

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The sum of two odd integers is even. Proof: If m and n are odd integers then there exists integers a,b such that m = 2a+1 and n = 2b+1. m + n = 2a+1+2b+1 = 2(a+b+1). Since (a+b+1) is an integer, m+n must be even.

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If n is a positive integer, then n is even iff 3n 2 +8 is even. Proof: We must show that n is even 3n 2 +8 is even, and that 3n 2 +8 is even n is even. First we will show if n is even, then 3n 2 +8 is even. n even means there exists integer a such that n = 2a. Then 3n 2 +8 = 3(2a) 2 + 8 = 12a 2 + 8 = 2(6a 2 + 4) which is even since (6a 2 + 4) is an integer.

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If n is a positive integer, then n is even iff 3n 2 +8 is even (cont.). Now we will show if 3n 2 +8 is even, then n is even using the contrapositive (indirect proof). Assume that n is odd, then we will show that 3n 2 +8 is odd. n odd means that there exists integer a such that n = 2a+1. 3n 2 +8 = 3(2a+1) 2 + 8 = 3(4a 2 + 4a + 1) + 8 = 2(6a 2 + 6a + 5) + 1, which is odd. Therefore, by the contrapositive if 3n 2 +8 is even, then n is even.

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n 2 + n + 1 is a prime number whenever n is a positive integer. Try some examples: n = 1, 1+1+1 = 3 is prime n = 2, 4+2+1 = 7 is prime n = 3, 9+3+1 = 13 is prime n = 4, 16+4+1 = 21 is not prime and is a counter example. Not true.

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n 2 + n + 1 is a prime number whenever n is a prime number. Try some examples: n = 1, 1+1+1 = 3 is prime n = 2, 4+2+1 = 7 is prime n = 3, 9+3+1 = 13 is prime n = 5, 25+5+1 = 31 is prime n = 7, 49+7+1 = 57 is not prime (19*3). Not true.

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Prove |x| + |y| |x + y| when x,y R. Note: |z| is equal to z if z 0 and equal to -z if z < 0 There are four cases x y 0 <0 0 0 <0 <0

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Prove |x| + |y| |x + y| when x,y R. Case 1 x,y are both 0 Then |x| + |y| = x + y = |x+y| since both x and y are positive.

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Prove |x| + |y| |x + y| when x,y R. Case 2 x < 0 and y 0 so |x| + |y| = -x + y If y -x, then x+y is nonnegative and |x+y| = x+y Since x is negative, -x > x, so that |x| + |y| = -x + y > x+y = |x+y| If y < -x, then |x+y| = -(x+y) = -x + -y. Since y 0, then y -y, so that |x| + |y| = -x + y -x + -y = |x+y|

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Prove |x| + |y| |x + y| when x,y R. Case 3 x 0 and y <0 so |x| + |y| = x + -y If x -y, then x+y is nonnegative and |x+y| = x+y Since y is negative, -y > y, so that |x| + |y| = x + -y > x+y = |x+y| If x < -y, then |x+y| = -(x+y) = -x + -y. Since x 0, then x -x, so that |x| + |y| = x + -y -x + -y = |x+y|

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Prove |x| + |y| |x + y| when x,y R. Case 4 x,y are both < 0 Then |x| + |y| = -x + - y = -(x+y) = |x+y| Therefore the theorem is true. This is know in mathematics as the Lipschitz condition.

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Prove that 3 3 is irrational. Proof (by contradiction): Assume that 3 3 is rational, i.e. that 3 3 = a/b for a,b Z and b 0. Since any fraction can be reduced until there are no common factors in the numerator and denominator, we can further assume that a and b have no common factors.

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Prove that 3 3 is irrational. (cont.) Then 3 = a 3 /b 3 which means that 3b 3 = a 3 so a 3 is divisible by 3. Lemma: When m is a positive integer, then if m 3 is divisible by 3, then m is divisible by 3. (Left as an exercise). By the lemma since a 3 is divisible by 3, then a is divisible by 3. Thus k Z a = 3k.

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Prove that 3 3 is irrational. (cont.) Now, we will show that b is divisible by 3. From before, a 3 /b 3 = 3 3b 3 = a 3 = (3k) 3. Dividing by 3 gives b 3 = 9k 3 = 3(3k 3 ). Therefore b 3 is divisible by 3 and from the Lemma, b is divisible by 3.

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Prove that 3 3 is irrational. (cont.) But, if a is divisible by 3 and b is divisible by 3, then they have a common factor of 3. This contradicts our assumption that our a/b has been reduced to have no common factors. Therefore 3 3 a/b for some a,b Z, b 0. Therefore 3 3 is irrational.

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Some Other Proof Strategies Rosen 3.1

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Backward Reasoning We have used mostly forward reasoning strategies up to now. However, sometimes it is unclear how to proceed from the initial assumptions. Backward reasoning may help-- Motto: If you can’t prove the original proposition, equate it to one you can prove.

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Prove (a,b Z +, a≠b)[(a+b)/2> ab] (i.e., the arithmetic mean is always greater than the geometric mean for this universe of discourse.) Backward reasoning proof (a+b)/2> abOriginal Assumption (a+b) 2 /4> abWhy? (a+b) 2 > 4abWhy? a 2 +2ab+b 2 > 4abWhy? a 2 -2ab+b 2 > 0Why? (a-b) 2 > 0Why?

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Prove (a,b Z +, a≠b)[(a+b)/2> ab] Backward reasoning proof (cont.) But, a,b Z +, (a-b) 2 > 0 implies a≠b. We can now easily start from a≠b and go backwards to reconstruct the path to prove the original proposition.

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Conjecture and Proof A conjecture is a plausible statement that has not been proved. A conjecture may result from recognizing that there are multiple examples for which it is true. For some conjectures, counterexamples are eventually found. However, even if a conjecture is valid for very many examples, this does not usually constitute a valid proof. (Why?)

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Conjecture and Proof Sometimes a complex proof is constructed as a series of conjectures that are then proved. Sometimes a proof is found by referring to the proof of a similar problem or class of problem. There are many famous conjectures that are still not proved (or only recently proved).

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Fermat’s Last Theorem x n + y n = z n has no solution for x,y,z,n Z, x,y,z≠0, n>2 This is a conjecture made by Pierre de Fermat (1601-1665), the French mathematician. He wrote in the margin of his copy of the works of Diophantus, an ancient Greek mathematician, that he had a “wondrous proof”, but that it wouldn’t fit in the margin. He then died, leaving no record of the proof!

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Fermat’s Last Theorem Attempts at proof over the years led to new fields, such as algebraic number theory. Finally, in 1994, Andrew Wiles provided a correct proof that required hundreds of pages of advanced mathematics.

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